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1:52 AM
@bolbteppa :I don't agree. I find the bundle picture of instantons using transition functions on the four sphere clearer than the physical interpretation using the "pure gauge at infinity"
The physical picture after you start thinking about it seriously often falls apart because some details are missing
4 hours later…
5:53 AM
Does anybody know a physical argument behind equivalence of dimensional regularization and momentum cutoff regularizations? I talked to a person who says to have faintly remembering about it in some talk by Howard Georgi where he talks of "shells" of mass being peeled off and idk...
1 hour later…
7:01 AM
@Sanjana See this
@bolbteppa do you mean for covariant vectors or defining connections in general? I'm more comfortable with the differential geometric definition of connections on vector bundles, but having said that I repeat that it's not about necessity
For me it's just that I don't like doing math intuitively. I like doing things rigorously, which doesn't mean that I need the most advanced proof available, just a self-consistent rigorous one. Then again, I think the difference is the same as speaking informal English vs higher register English
You wouldn't speak informally to the PoTUS in a formal meeting, would you? :P
7:37 AM
I have read about a dozen pieces of text on gauge transformations, diffeomorphisms and co-ordinate transformations... it's all gone over my head. does anyone know if there's a gentle introduction somewhere that relates the different usages in different contexts (EM, GR, cosmology)?
8:27 AM
It's not clear to me whether or not Carroll is in disagreement or not with e.g. Wald on this topic...
9:01 AM
anyone want to play this? lichess.org/G9bmSkd0
9:20 AM
@qwerty r u interested in qft and gr?
@RyderRude more just gr, cosmology, em.
not quite up to speed with qm/qft in general.
Carroll says "you have to cheat somewhat by equating components of tensors in two different coordinate systems" to treat gauge transformations as coordinate transformations and not diffeos. wald says the active/ passive viewpoints (diffeos v coordinate transformations) are really equivalent.
9:36 AM
@qwerty oh
Hello Everyone...
Hello @RyderRude
@123 hello
@qwerty i guess the outcome of a diffeomorphism is indistinguishable from that of a co ordinate change. if u take a co ordinate change, and interpret the co ordinates as the "literal points" instead of labels of points, u end up interpreting the co ordinate change as an active diffeomorphism
but there may be subtle differences.. maybe they can be ignored
@RyderRude i don't understand the intuition behind gradient, divergence and curl. I am thinking about these since 4 years. But still no progress. I have learned many threads and books but no clue i found.
First pls help me in understanding divergence.
@RyderRude oops sorry new to this chat thing. yes i suppose that is the point that Wald makes in his textbook. yet e.g. bertschinger claims things like the volume element in GR is invariant under diffeos, but not coordinate transformations; so they cant be treated the same.
@123 consider a fluid, draw a closed surface as you like. the divergence tells you how much stuff is entering or leaving the surface
9:52 AM
@qwerty Hello yes i have read this intuition many times. But the problem is that i can not connect this intuition with mathematics.
@qwerty do u mean, like, $d^4 x$ is a tensor density when interpreted in terms of co ordinates, but a tensor when interpreted as a differential form?
I want to understand this step by step. Like we have vector field (e.g. gravitational field, electric field) then if we partial differentiate this field at some point which gives slope/gradient of this field keeping other variables constant.
@RyderRude bottom of page 9 web.mit.edu/edbert/GR/gr5.pdf
So the dot product between partial differentiation and vector field gives specific slope in that direction.
I want such kind of description.
@123 look up the divergence theorem
9:58 AM
@RyderRude Ooookay.. let me read again
10:30 AM
@qwerty This topic is a horrible can of worms and really people do not seem to agree with each other. Personally I'm on Wald's and the mathematicians' side that you really only need to talk about diffeomorphisms and the active/passive distinction is a trap you don't need to fall into.
@ACuriousMind is there any intro that you might recommend in particular? the more i read the more confused i get; i feel like the fact that people often start talking about gauge transformations without defining what they mean (and it seems there are a few related meanings? at least when people say "gauge invariant") only adds to the confusion.
yes, that's true; I don't think there is a single source that does it perfectly
I think the best work on gauge theory that's consistent from start to finish is Quantization of Gauge Systems by Henneaux and Bunster, but it's almost entirely in the Hamiltonian formalism and you need to do a bit of work to connect it to what other people are doing
alas, nothing will solve the problem that people use words inconsistently - as you say, people often try to talk about "gauge transformations" only handwavingly defining them as some kind of "spacetime-dependent transformations" when you really need a more solid definition to be able to work with it
why do you think that people don't seem to agree with each other? is it inconsistent definitions? i somehow dont think the definitions are inconsistent (at least when talking about specifically cosmology)
also thanks for the book rec. i'll give it a look.
@qwerty It's just people not being careful enough about what exactly they mean mathematically when they say things :P
10:48 AM
hmmm i'm inclined to agree in general!! but in this case both carroll and bertschinger seemed to be precise in what they were saying with respect to diffeos/coordinates, or at least as much as wald; even though i'm still confused.
Everyone is I fear
It is a long history of it :p
@qwerty Well, the first problem here is: Could you write down what exactly the definition of "diffeomorphism" and "coordinate transformation" is that each of these sources are using? Like, a proper, rigorous definition, not just a handwaving sentence that sounds like a definition?
@Slereah "8 decades of dispute" eep
14 hours ago, by ACuriousMind
at this point I feel like I'm linking this once per week but I explain this in a bit more detail (in particular footnote 2 and the expansion of $F_\omega$) in this answer of mine
@qwerty Longer even really
10:51 AM
14 hours ago, by ACuriousMind
that answer is the result of me fighting with the sentence "GR is a gauge theory" for several years :P
It's just a continuation of the long debate about relative or absolute space
Don't feel too bad about not getting general covariance fully, it is a pretty thorny topic and it's not super important in most contexts
@ACuriousMind hah... yeap, and i asked something related last year here physics.stackexchange.com/q/763220/92181
i never understood the answer that was posted...
There's probably like half a dozen people define it
*half a dozen ways
i see... Carroll mentioned in appendix b of his textbook actually, but how gauge invariance is related(ish; i dont know if he agrees - he says it is "garbled") to general covariance in appendix b of his textbook...
The most important part of general covariance to know is mostly how it affects the constraints in GR
11:06 AM
@ACuriousMind hm, point taken. personally if any source pinned down what gauge transformation really is i would be really happy.
11:37 AM
@qwerty in standard model, gauge fields arise by making global internal symmetries into local and writing the covariant derivative, and then writing the free term of the gauge field using its curvature
so this is a unified view of gauge symmetries other than GR
a common element of guage fields is that theyre connections
you don't "make global symmetries into local ones"
@qwerty The answer is basically saying that gauge transformations in GR are coordinate transformations
Because the exact same physical setup can be described using different coordinate systems
The second part is saying that coordinate transformations are equivalent to diffeomorphisms (actually isometries in proper mathematical language)
11:54 AM
@ACuriousMind yeah.. i was just saying that this trick is how gauge fields of the known forces happen to arise, but we dont have a good explanation for why this trick should work
12:26 PM
If the Maxwell tensor in EM is the analogue of the Riemann curvature tensor in GR, is $A_\mu$ the Levi-Civita connection, or some other choice of connection? And if not, is there some reason why we don't use the L-C connection in EM (like it doesn't exist, or we just don't introduce a metric in E&M)?
@Charlie the $A_\mu$ is a connection on a U(1)-principal bundle, not on the frame bundle like the L-C connection
Ahh yeah
Another question I was thinking about this morning, when we mention the connection coefficients $\Gamma$ in GR we have to be careful about the non-tensorial transformation rules and emphasising that two of the components label the Lie algebra element (matrix) components, but if the Riemann curvature tensor is also a Lie algebra-valued 2-form it also has two indices which denote the indices on a Lie algebra element, but it doesn't have the same transformation property and we treat it as a
completely ordinary tensor, what gives?
It's part of the reason I'd never connected the Riemann CT and the connection 2-form, because I've always been taught it's just a regular tensor
recall that also in generic gauge theory, while the $A$ transforms non-linearly under gauge transformations: $A\mapsto g^{-1}Ag + g^{-1}\mathrm{d}g$, the curvature transforms linearly: $F\mapsto g^{-1}Fg$
Though I guess it's definition does make use of the $Gamma$ which are weird
in the same way the Christoffels are non-tensorial while the Riemann tensor is tensorial
12:32 PM
Hmm oh yeah
1:23 PM
i feel the matter field is also technically a gauge field
because it has just as many redundancies. it also transforms under the gauge transform
so if we define "guage field" to mean "redundant field", we hav to include these matter fields
but one crucial distinction is that this redundancy is not there in the free theory of the matter field, while it's still there in the free theory of the EM field
but technically, the redundancy is still there in the free matter field theory too. it's just that the curvature of the connection is zero
but when we write the free field solution of the matter field in qft, we have already chosen the trivial gauge. so no more degrees of freedom need to be removed
1:41 PM
@SillyGoose it is far more likely that a theoretical physicist cranked out a few computations and then gets bothered by some regularities and then sets out to discover that they are a single result. Mathematical physicists are much less likely to actually be involved in a computation of that form.
2 hours later…
3:32 PM
When you contract the Riemann tensor to get to the Ricci tensor/scalar, are you summing over the Lie algebra indices or the 2-form indices?
The Christoffel symbols don't have Lie algebra indices, that's the spin connection :p
I'm not sure what you mean, unless it's a technicality
@Slereah they do
they have $\mathfrak{gl}(n)$ indices
@Charlie that's the tricky part - it's a mixture, which is why there is no analogue of the Ricci tensor in generic gauge theories
I see
The Ricci tensor and scalar are genuinely specific to GR
3:38 PM
I'm still recovering mentally from this chat.stackexchange.com/transcript/message/65529747#65529747
Oh it doesn't put the comment up
it does if you only post the link
Ah right
19 hours ago, by ACuriousMind
The Riemann tensor is the curvature form
This bridged a gap between two topics that I've separately learned in a really satisfying way
What a time to be alive
homology considers maps from standard r-simplices to the manifold and makes it a group, and homotopy considers maps from n-cubes to the manifold with boundary identified, and makes it a group
what is the generalisation of these ideas
why consider maps from those two spaces
is there a general idea here or do these maps just happen to be useful
4:27 PM
what are some ways to write mixed partial derivatives as an operator
like $\partial_{xz}$ or $\frac{\partial}{\partial x}\frac{\partial}{\partial z}$ or $\frac{\partial}{\partial_{xz}}$?
or maybe even $\frac{\partial^2}{\partial x\partial z}$ or $\frac{\partial^2}{\partial_x\partial_z}$?
4:40 PM
There are all kinds of notations, the most common notation is certainly $\frac{\partial^2}{\partial x\partial z}$ but the first two examples you've given probably make the most sense to the average person who knows what a partial derivative is
I've never seen your last example, since it doesn't suppress anything beyond $\frac{\partial^2}{\partial x\partial z}$, it's just more keystrokes to write basically the same thing
ok ill do the most common one then thank u
what would I call this quantity
$$\nabla\times\nabla = \hat{i}\left[\frac{\partial^2}{\partial z\partial y}-\frac{\partial^2}{\partial y\partial z}\right]-\hat{j}\left[\frac{\partial^2}{\partial z\partial x}-\frac{\partial^2}{\partial x\partial z} \right]+\hat{k}\left[\frac{\partial^2}{\partial y\partial x}-\frac{\partial^2}{\partial x\partial y} \right]$$
we have $\nabla^2 = \nabla\cdot\nabla$ the laplacian or laplace operator
5:07 PM
i guess if mixed partials are equal this is the zero operator nvm math.stackexchange.com/questions/1570280/…
It's the statement that the curl of the gradient is zero
This answer is confusing me. AFT gives an argument for $\Delta\sim \frac{s^J}{s-m^2}$
thought this comment was pretty cool
The formula on GSW has $\frac{s^J}{t-m^2}$ though
@Obliv yeah, you'd have to break continuity of second derivatives to have it different from $0$
For example, if you apply that operator to $f(x,y)=\tan^{-1}(y/x)$
Of course if you stay outside the origin it is zero
If you want to consider the origin though it yields something distributional $\sim\delta^{(2)}(\vec{x})$
5:24 PM
i've heard that term distributional before, when talking about fourier transform of $e^x$
not sure what it means :o
@Obliv in this case I mean a Dirac delta :P
oh ok
If you wonder why I had that example ready, vortices in the XY model :P
well I thank u for that :P
well this was a waste of time because in the problem statement they forgot the parentheses in this expression $\nabla\times\nabla\times\vec{A} = -\nabla^2\vec{A}+\nabla(\nabla\cdot\vec{A})$
should be $\nabla\times(\nabla\times\vec{A})=-\nabla^2\vec{A}+\nabla(\nabla\cdot\vec{A})$
Oh @bolbteppa that question is yours. Did you find anything?
6:09 PM
@Charlie this derivation of the Riemann tensor shows how to see the Riemann tensor as a 2-form with two additional indices naturally, and combining with my post on connections yesterday, you can see that if $\mathbf{A} = A^{\rho} \mathbf{e}_{\rho}$ lives in the same space as your $x^{\mu}$, then it has all space-time indices,
however if you take your vector (the $\mathbf{A}$ now denoted) $ \mathbf{V}$ to live in some other space say with $SU(N)$ indices, you'd get Yang-Mills curvature with the connection (now denoted $\mathbf{A}$ as usual for YM connections) as I explained yesterday
@Mr.Feynman I found a complicated way to do it and I haven't written it up nicely yet
6:27 PM
Can someone help me understand this statement:
A 1st order phase transition is therefore a transition where the first derivative (dG/dT) is discontinuous, which is equivalent to a latent heat at the phase transition.
If the system has a 2nd order phase transition, does that mean that the latent heat is equal to the 2nd derivative w.r.t Temperature of Gibs free energy ?
7:15 PM
@bolbteppa damn, why do GSW write it like it's trivial?!
Wait, are you talking about the general form? Do you have at least a shortcut for the asymptotic form?
@ACuriousMind i started skimming through "Quantization of gauge systems", and it's quite helpful, thank you again! although now im wondering why some of the introductory content isnt covered in classical mechanics texts like goldstein...
I think it's because in the particle mechanics context you can usually get rid of first-class constraints by fully gauge-fixing them and then get rid of all the second-class constraints by passing to a reduced phase space; if that works you can just start with the reduced description without having to develop the theory of constraints at all
7:33 PM
I just finished my phd and I'm finally feeling ready to understand undergraduate physics...
@Mr.Feynman if you find a simple straightforward way to show it I would love to see it
7:55 PM
@bolbteppa I wish, I'm just feeling a little bit flexed on the second page of this book :P
Just skim the first chapter at best, start from the second one
@qwerty 2014 ACM: "I just finished my undegrad and I'm finally feeling ready to understand phd physics..."
The point of the first chapter is to show you there's around 50 pages of motivation and background all based on insane stuff like this formula for the basic models they set up from first principles in the second chapter on
@bolbteppa yeah, I guess I will. Each page of chapter 1 is supposed to be years of research of the 60s
8:24 PM
@ACuriousMind and i think it doesnt directly talk about EM because it's a gauge transformation "of the second kind"? unless i missed it
@qwerty What doesn't talk about EM?
oh sorry. the quantization of gauge systems book
one of the last chapters is about EM
they call it "the free Maxwell field"
the main weakness of the book is that they don't do examples outside of the exercises and those last chapters
ah yeap i see it now
8:54 PM
What a surge in recent talk about gauge theory xD
@RyderRude What do you mean by your first statement on homology. To my understanding homology considers a functor which maps from one category into the category of chain complexes, then you talk about n-cycles that are not n-boundaries
And the space of n-cycles that are not n-boundaries is the nth homology class and through some theorems and etc tell you something about the space you functored into a chain complex
9:20 PM
guys how come this equation adds a scalar to a vector on the right side ?
I don't see how we can do this
@Obliv it doesn't
doesn't the laplacian return a scalar? $\vec{\nabla}^2\vec{A} = \nabla\cdot(\nabla\vec{A})$
what do you think $\nabla \vec A$ is?
"the Laplacian" acts on functions, not vectors, the $\nabla^2 A$ is the "vector Laplacian" where you just apply the Laplacian to every component of the vector
so this equality is meant for scalar/vector fields, not individual vectors?
I didn't formally learn the laplacian so I just assumed it acted on both
I don't understand the question
you can't differentiate a vector, you can only differentiate a vector field
my point is just that $\nabla^2 A$ is still a vector (field)
in Cartesian components $(\nabla^2 A)_\mu = \partial_\nu \partial^\nu A_\mu$
9:34 PM
WAIT so it's not a scalar AT ALL?
Do u gusy rhink enm will be taught with differential forms in the future
er no, it turns $\vec{A}$ into a scalar field by $\nabla^2\vec{A}$
@SillyGoose why would it? differential geometry isn't that new
I'm so confused, because I explicitly asked my prof if laplacian returns a scalar and he said yes
so if it's applied to a vector field it returns a vector field but if it's applied to a scalar field it returns a scalar field?
again, it's not "the Laplacian"
just because the symbol $\nabla^2$ is the same it doesn't mean it's exactly the same operator
"the Laplacian" applied to a scalar field returns a scalar field
but you can also consider the "vector Laplacian" applied to a vector field, which returns a vector field where each component is the scalar Laplacian applied to the component of the original vector field
9:41 PM
but isn't it the div. of the gradient of a ___ field. which will be scalar regardless since div. gives a scalar?
you can't take the "div" of anything but a vector field
there's no "grad" of a vector field, either
i think ive only ever seen the scalar laplace operator definition which is where the confusion is coming from
the vector definition is completely different
not sure why it's denoted with the same notation...
what do you mean, "completely different"? In Cartesian coordinates, it's literally the scalar Laplacian applied to each component of the vector field
didn't see that :P but yeah I think the definition above that one is something I should prove first before I am to prove the original problem
otherwise i can just simply plug that in
makes it pretty trivial
10:04 PM
is it standard to denote $\vec{A}$ as $(A_x,A_y,A_z)$ or whatever components we choose? Do we reserve $\langle \rangle$ for individual vectors?
nvm i think we used parentheses for fields
once again, I do not understand the question :P
like, if $\vec{A}$ is a vector field, for component notation can we write $(A_x,A_y,A_z)$?
well, im gonna do the smart thing and just explicitly say what i'm talking about rather than hope the reader understands my conventions/notation
10:20 PM
@ACuriousMind well stuff like density operator formalism of QM isn't new either but still not in textbooks on UG or grad quantu
sans a single section or two
@Obliv you can do whatever you would like. the $\langle - , - \rangle$ notation sucks in my opinion because this symbol is reserved for inner products in my head
yeah I wanna reserve that notation for other stuff too
i think in physics it is common to write $\vec{A} = (A_1, A_2, ..., A_n)$ and denote the $i^{th}$ component by $A_i$
if you are talking about a vector whose components correspond to $x, y, z$ directions, just define that $1 \mapsto x$, $2 \mapsto y$, and $3 \mapsto z$
I wish I reserved $A_i$ to mean $\frac{\partial \textbf{A}}{\partial i}$ though..
actually no
well maybe i should set up the notation i mentioned more carefully.
cuz then I can stack the notation like so $\partial^kA_i$ to mean $k$th partial of $i$ component of $\textbf{A}$
10:27 PM
right that is the usual notation i believe
define $\partial_i$ as the partial derivative with respect to $x_i$ and define $A_i$ as the ith component of a vector/ vector field
and in cartesian coordinates, $x_1 \mapsto x, x_2 \mapsto y, x_3 \mapsto z$ but in general you can just have some arbitrary coordinates $x_i$
ohh so thats what ACM meant by the notation he used earlier
@SillyGoose well, you have to be careful, there's a reason I said Cartesian several times above
$\partial_{\nu}\partial^{\nu}$ to mean $\nu$-th partial of $\nu$-th component
if your coordinates are not Cartesian, then the Laplacian (either scalar or vector) is not $\partial_\mu \partial^\mu$, but includes terms incurred from non-zero Christoffels
wait but then u used $A_{\mu}$
10:30 PM
oh i see. i didn't mean to talk about the laplacian or anything just setting up notation
@Obliv are you familiar with einstein summation notation
who is einstein?
does he play in the NFL
ok I know we already discussed $\int dx (\dots)$ vs $\int (\dots)dx$ notation but what about $\hat{i}\left[blah\right]$ vs $\left[blah\right]\hat{i}$ notation
I kinda like the first one simply because I think it's more readable
heuristically: since you're doing electromagnetism, which is a relativistic theory, we will have four coordinates: one coordinate for time $x^0$ and three coordinates for space $x^i$ where $i = 1, 2, 3$. We combine the $0$ index and the $1,2,3$ indices into a single free index denoted by a greek index. This is for instance $\mu$ as ACM writes. Then, if I have an expression $A^\mu A_\mu$ in which the indices are repeated (and also on different floors), then this expression is short hand for
$\sum_{\mu = 0}^3 A^\mu A_\mu$
where we are writing $\vec{A} = (A^0, A^1, A^2, A^3)$ in terms of its components as $A^\mu$
@Obliv so this is not the $\nu$th partial of the $\nu$th component, it is the second partial derivative w.r.t. $x_\nu$
so if $\textbf{A} = (A^0,A^1,...) = \sum\limits_{\mu=0}^3 A^{\mu}$ then what is $A_{\mu}$
no the summation is only implied when you have two repeated indices that are the same
$\vec{A} \mapsto A^\mu$ in components, but there is just one $\mu$ index
$A^\mu$ is literally the $\mu$th component of $\vec{A}$
are you familiar with a vector space $V$, thinking of the dual to the vector space $V^*$ as the space of linear functionals over $V$?
I understood maybe 30% of those words
10:43 PM
also the wiki is probably more organized and accurate than i am at presenting the einstein summation convention: en.wikipedia.org/wiki/Einstein_notation
also im not sure how we got here it seems a little divorced from your original question :P
Yeah I'm not familiar with multilinear algebra or tensor calculus so I don't understand a lot of what's written here
well the idea is not complicated. you start with a vector space $V$. it has elements $v \in V$. consider the set of linear maps $\omega: V \to \mathbb{R}$. we given this map a name: a linear (because it is...linear) functional (because it is a function of vectors).
the set of all such $\omega$ turns out to naturally have the structure of a vector space itself, and we call it the dual space $V^*$ to $V$
what's a linear map? Does it just mean it's injective
it means it respects the structure of a vector space
so like a homomorphism?
10:48 PM
$f(\alpha\vec{x} + \beta \vec{y}) = \alpha f(\vec{X}) + \beta f(\vec{y})$
yes linear maps are precisely the homomorphisms for vector spaces
cool that is pretty simple then
this is also roughly the general definition of what it means to be a homomorphism of a mathematical structure: you respect or preserve the structure
I wonder if $\mathbb{R}^*$ meaning the group of the nonzero reals under mult. and $\mathbb{R}^{**}$ are duals. yeah probably not since $\mathbb{R}$ isn't a group by itself under mult.
just recycled symbols most likely
well the construction i just talked about is for vector spaces not groups
so is a dual space a space that is structurally similar
10:52 PM
you can probably generalize this notion that i described but im not sure how well-defined it would be in general
i am just talking about vector spaces and what the dual space to a given vector space is
i am not claiming a general definition of what a dual space is
i did claim a general notion of a homomorphism but that is all
so an example could be $\mathbb{R}^n$ is the dual space to $\mathbb{R}$
oh wait linear functional means the set of all linear maps ok
for at least vector spaces over $\mathbb{R}$ the dual space has the same dimension as the original space
if you want to think concretely
think of $V = \mathbb{R}^n$ and its elements are column $n$-vectors with real entries
then $V^*$ is the space of row $n$-vectors with real entries
:65534547 linear already means that it preserves vector space structure
ok I think I see
10:56 PM
it needs to be linear and take in vectors as input and spits out real numbers (more generally elements of the field $V$ is defined over)
this is relevant because $A^\mu(x^\mu)$ at each point $x^\mu$ lives in a vector space $V$ while $A_\mu(x^\mu)$ at each point $x^\mu$ lives in the space dual to $V$, i.e. $V^*$
so then it actually makes sense what $A^\mu A_\mu$ is.
this is at least how i think about it but other people are more familiar with this sort of stuff :P.
is this kinda similar to transformations?
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